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How to represent an inductive circuit

  1. Jan 19, 2012 #1
    In all the tutorials I have seen on inductance and magnetism, I have never seen how to represent an inductor as a voltage or current source in a circuit. Please see figure. I am trying to determine V and I given the characteristics of the coil and the magnetic flux through the coil.

    Note: Before you suggest an answer, would you please make sure your solution works for all values of resistance from zero to infinity? For example, Faraday's law for induced voltage only works into an open circuit. Thanks for suggestions.

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    Last edited: Jan 19, 2012
  2. jcsd
  3. Jan 19, 2012 #2


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    Staff: Mentor

    Welcome to the PF.

    Don't see an attachment yet, though...

    EDIT -- Attachment added now.
    Last edited: Jan 19, 2012
  4. Jan 19, 2012 #3
    what do you mean when you say 'Faraday's law for induced voltage only works into an open circuit'?
  5. Jan 19, 2012 #4
    I just added the schematic to my OP. To technician, if you have a solution that uses Faraday's law, by all means post it. I would like to focus on the solution to my problem. Thanks!
  6. Jan 19, 2012 #5
    If you need an expression for the emf produced by the coil it comes from Faradays law....
    induced emf = rate of change of magnetic flux linkage. (magnetic flux linkage = flux x number of turns on the coil = N∅)
    Therefore e = -d(N∅)/dt
    Hope this helps with the rest of your analysis.
  7. Jan 19, 2012 #6
    Sorry, but it doesn't help and I hope there are other solutions out there. If you have a coil with two terminals and you short them together, clearly you can't have infinite current, so how does this help me? I tried to make this clear in my OP.
  8. Jan 19, 2012 #7
    Sorry to hear that the responses you have had have been of little help.
    Here is some more explanation:
    If you have a coil experiencing a changing magnetic flux linkage there will be an induced emf
    given by e = -d(N∅)/dt (as previously stated)
    If you connect the ends of the coil together (assuming there is zero resistance) then a current will flow (it will not be infinite but....) the current will increase according to Faraday's law...
    e = -LdI/dt (L is the inductance of the coil) so you have an expression for the rate of rise of the current. The current will increase uniformely (towards infinity) at a rate = e/L.
    If there is resistance in the circuit then there will be a max value for the current (= e/R)
    I hope that this information can be of some help in understanding what you need to know
  9. Jan 19, 2012 #8
    You don't see it because it's not part of circuit theory.

    There's no such thing as a magnetic field in circuit theory.
  10. Jan 19, 2012 #9
    I agree, although magnetic field theory does give values for emf.
    Without a changing field in this example there would be no emf and therefore no problem!
  11. Jan 19, 2012 #10
    My guess is the solution lies in this formula

    I = VS / (R + XL)

    I wish I could give a dollar to the first person who can draw the circuit correctly and give me a complete solution. It may be that VS is a voltage source that is a function of webers.
  12. Jan 19, 2012 #11
    I may not be able to draw the circuit you are after but it will not have I =Vs/(R+Xl)
    The combination of R and Xl in series is called impedance....Z....and the relationship between R and Xl is
    Z^2 = R^2 + Xl^2
    and I = V/Z
    I wish that I could grasp what you are after!!
    Your last post suggests that you are dealing with an AC circuit problem
  13. Jan 19, 2012 #12
    Yes, but that's physics. In circuit theory the definition of the inductor is V/L = di/dt.
  14. Jan 19, 2012 #13
    Yeah, that's how circuit analysis is done - you have real and imaginary components, or resistive and reactive. That's why (R + XL) is correct. You don't have to square them.

    I really appreciate all of the comments I have received. At this point can we throw this open to other electronics fans out there? Just to review, we need a circuit that allows us to calculate current in the resistor and works for any value of resistance. We also know that the voltage or current generator will only produce AC signals and its amplitude will be related to flux.

  15. Jan 20, 2012 #14
    I see what you mean now.... you mean Z = R + jωL I think
    and Z = R -jωC for capacitance.
  16. Jan 22, 2012 #15
    The inductive coil by itself does not act as a voltage source, just a voltage drop, like any other impedence to current. It becomes a voltage source when it is the secondary coil in a transformer, and then of course only to an AC current. Calculations for current are according to Ohm's Law, for simple electronic AC circuits, which I presume is what you're dealing with.
  17. Jan 29, 2012 #16
    I think Mr. Faraday would be very surprised to hear this. Think about the secondary of a transformer. How does that coil know where the flux is coming from? It just has flux in it. So why is that any different from any other coil in a magnetic field? Are you saying the only source of a magnetic field is the primary of a transformer? By the way, the spelling is "impedance".

    I have asked the moderators to cancel this thread as it is 10 days old and we are nowhere getting an answer to my question and I don't think anybody is going to read through all of this. Just my opinion.
  18. Jan 29, 2012 #17
    Perhaps that is because your question is unclear.

    An inductor is not, of itself, a voltage or current source.

    If you connect one into a circuit you do not magically get either source included in your circuit.

    So an inductor is an inductor, a voltage source is a voltage source and a current source is a current source.
    Each is different and thus awarded a different circuit symbol and obeys different circuit equations.

    So if you would like to try to explain your question again more clearly we could respond more accurately.
  19. Jan 29, 2012 #18
    Yes, studiot is correct. Your question is unclear.

    Mr. Faraday knew nothing of circuit theory. He was doing physics.

    In circuit theory the inductors satisfy simple differential equations on their terminals. There are no fields and no generators. To model general coupling between multiple coils you have mutual inductance. You still don't have magnetic fields or sources in this case, just coupling between the windings.

    Your confusion seems to stem from a mixing of circuit and field concepts.
  20. Jan 29, 2012 #19

    jim hardy

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    induction works quite well.

    here let's look at the physical process....

    if you place an open circuited coil in a region of changing magnetic flux
    it will have a voltage induced in it, n*dphi/dt.

    if you place same coil, but now with its ends shorted, in same region
    current will flow in that coil because the induced voltage is impressed across the resistance of the coil.
    Said current causes a MMF that opposes the changing flux , this is Lenz's Law.
    So the current in the coil will be enough to mostly cancel out the changing flux.
    Just enough flux change remains to make required voltage for IR drop of coil.
    IF the resistance of the coil were zero it'd perfectly cancel out all the changing flux.

    This is not pie-in-sky theory, it's exactly how a current transformer works.

    A voltage transformer works similar, except when the load current in secondary starts to cancel flux, more current flows in primary to push flux back up. So primary current follows secondary.

    It's a shame we dont have access anymore to the 1880's magazines where they were figuring this stuff out. They give beautifully logical explanations and we see how the pioneers struggled with same concepts. The voltage transformer concept above was described around 1880 in engineering journals as "...beautifully self regulating".

    So - OP's question would be answered by observing that phi is total flux - sum of whatever original flux links his coil AND the flux created by current flowing in the inductance of his coil.
    after all - What is inductance? Flux linkages per amp.

    d(phi) is all the flux, not just part of it.

    So in OOP's diagram write e= n * d(phi1 - phi2) where phi2 is Li/n represented by an arrow pointing up.
    Last edited: Jan 29, 2012
  21. Jan 30, 2012 #20
    I think the confusion arises from the original picture posted by jwriter and the thread title. (no offence meant).

    This is because the picture shows magnetic flux.

    What magnetic flux? Where does it come from?

    It should be noted that there are two kinds of inductance. Self inductance and mutual inductance. It is not clear which we are dealing with here.

    This is important because current/voltage sources are sources of energy input to a circuit.
    As previously noted, by itself an inductor is not an energy source.
    The flux forms a source of energy.
    Combined with an inductor this can input energy to a circuit, but the governing equations are different from those of a current/voltage source. These equations have already been mentioned by others.

    Just as a matter of interest is this anything to do with Professor Lewin's famous lecture 'Kirchoff v Faraday' ?
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